\[x^2+(y-1)^2=4\]

**In the manner describe:****a) One around clockwise starting at $(2,1)**$**b) Three times around counterclockwise starting at $(2,1)$**

This question **aims** to understand the **parametric equations** and **dependent** and **independent** variables concepts.

A sort of equation that uses an **independent** variable named a **parameter** (t) and in which **dependent** variables are described as **continuous** functions of the parameter and are not **dependent** on another existent **variable.** When necessary More than one **parameter** can be used.

## Expert Answer

Given that a **particle** moves around the circle having **equation** is $x^2+(y-1)^2=4$.

**Part a:**

$x^2+(y-1)^2=4$ is the path of the **circle** in which the particle moves in the manner once **around clockwise,** starting at $(2,1)$

\[x^2+(y-1)^2=4\]

\[\dfrac{x^2}{4}+\dfrac{(y-1)^2}{4}=1\]

\[\left(\dfrac{x}{2}\right)^2+\left(\dfrac{(y-1)}{2}\right)^2=1\]

$\cos^2t + \sin^2t =1$ is the **parametric equation** of the circle.

As the circle is **revolving** once in the **clockwise** direction then the limit $t$ is $0 \leq t \leq 2\pi$

By comparing the two **equations** $\left(\dfrac{x}{2}\right)^2 +\left(\dfrac{(y-1)}{2}\right)^2 =1$and$\cos^2t +\sin^2t=1$.

\[\dfrac{x}{2}=\cos t\space\space and \space\space\dfrac{y-1}{2}=\sin t\]

\[x=2\cos t\space\space and\space\space y-1=2\sin t\]

\[x=2\cos t \space\space and\space\space y=1+2\sin t \space\space \epsilon\space |0, 2\pi|\]

**Part b:**

$x^2+(y-1)^2 =4$ is the **path** of the circle in which the **particle** moves in the manner three **times** around **counter-clockwise,** starting at $(2,1)$

\[x^2+(y-1)^2=4\]

The **circle** has a radius of $2$ and the **center** is at $(0,1)$.

As the circle is **revolving** thrice, the $t$ is less than **equal** to $3(2\pi)$ that is, $0\leq t\leq 6\pi$

By **comparing** the two equations $\left(\dfrac{x}{2}\right)^2+\left(\dfrac{(y-1)}{2}\right)^2=1$ and $\cos^2t+\sin^2t=1$.

\[\dfrac{x}{2}=\cos t\space\space and \space\space\dfrac{y-1}{2} =\sin t\]

\[x =2\cos t\space\space and \space \space y-1= 2\sin t\]

\[x =2\cos t\space\space and \space \space y=1+2\sin t \space\space\epsilon\space |0, 6\pi| \]

## Numerical Answer

**Part a**: $ x = 2\cos t \space \space and \space \space y = 1+2\sin t \space \space \epsilon \space |0, 2\pi| $

**Part b:** $ x = 2\cos t \space \space and \space \space y = 1+2\sin tÂ \space \space \epsilon \space |0, 6\pi| $

## Example

A **particle** moves along the circle. Find its **parametric** equation for the path in the **manner** halfway around **counterclockwise** starting at $(0,3)$.

$x^2 + (y-1)^2 =4$ is the path of the **circle** in which the particle moves in the **manner** halfway around **counter-clockwise,** starting at $(0,3)$.

\[x^2 + (y-1)^2 =4 \]

point $(0,3)$ lies on the y-axis.

\[\dfrac{x^2}{4} + \dfrac{(y-1)^2}{4} =1 \]

\[ \left( \dfrac{x}{2} \right)^2 + \left( \dfrac{(y-1)}{2} \right)^2 =1 \]

$\cos^2t + \sin^2t =1$ is the parametric equation of the circle.

As the **circle** is revolving in halfway around the **counterclockwise** direction, the **limit** $t$ is $\dfrac{\pi}{2} \leq t \leq \dfrac{\pi}{2} + \pi$

That is: $\dfrac{\pi}{2}\leq t \leq \dfrac{3\pi}{2}$

By **comparing** the two equations $\left( \dfrac{x}{2} \right)^2 + \left( \dfrac{(y-1)}{2} \right)^2 =1$ and $\cos^2t + \sin^2t =1$.

\[ \dfrac{x}{2} = \cos t \space \space and \space \space \dfrac{y-1}{2} = \sin t \]

\[ x = 2\cos t \space \space and \space \space y-1 = 2\sin t \]

\[ x = 2\cos t \space \space and \space \space y = 1+2\sin t \space \space \epsilon \space |\dfrac{\pi}{2}, \dfrac{3 \pi}{2}| \]